f(x)=-3x^2+6x-2 eching

$$f ( x ) = - 3 x ^ { 2 } + 6 x - 2$$

Omil

$$-3\left(x-\left(-\frac{\sqrt{3}}{3}+1\right)\right)\left(x-\left(\frac{\sqrt{3}}{3}+1\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$-3x^{2}+6x-2=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\left(-2\right)}}{2\left(-3\right)}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\left(-2\right)}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+12\left(-2\right)}}{2\left(-3\right)}$$

12 ni -2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-24}}{2\left(-3\right)}$$

36 ni -24 ga qo'shish.

$$x=\frac{-6±\sqrt{12}}{2\left(-3\right)}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{3}}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{3}}{-6}$$

x=\frac{-6±2\sqrt{3}}{-6} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}-6}{-6}$$

-6+2\sqrt{3} ni -6 ga bo'lish.

$$x=-\frac{\sqrt{3}}{3}+1$$

x=\frac{-6±2\sqrt{3}}{-6} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{3} ni ayirish.

$$x=\frac{-2\sqrt{3}-6}{-6}$$

-6-2\sqrt{3} ni -6 ga bo'lish.

$$x=\frac{\sqrt{3}}{3}+1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 1-\frac{\sqrt{3}}{3} ga va x_{2} uchun 1+\frac{\sqrt{3}}{3} ga bo‘ling.

$$-3x^{2}+6x-2=-3\left(x-\left(-\frac{\sqrt{3}}{3}+1\right)\right)\left(x-\left(\frac{\sqrt{3}}{3}+1\right)\right)$$

Baholash

$$-3x^{2}+6x-2$$